Answer in the attachment.
ANSWER
B.Yes, f is continuous on [1, 7] and differentiable on (1, 7).
![c\approx 3.08](https://tex.z-dn.net/?f=c%5Capprox%203.08)
EXPLANATION
The given
![f(x) = ln(x)](https://tex.z-dn.net/?f=f%28x%29%20%3D%20ln%28x%29%20)
The hypotheses are
1. The function is continuous on [1, 7].
2. The function is differentiable on (1, 7).
3. There is a c, such that:
![f'(c) = \frac{f(7) - f(1)}{7 - 1}](https://tex.z-dn.net/?f=f%27%28c%29%20%3D%20%5Cfrac%7Bf%287%29%20-%20f%281%29%7D%7B7%20-%201%7D%20)
![f'(x) = \frac{1}{x}](https://tex.z-dn.net/?f=f%27%28x%29%20%3D%20%5Cfrac%7B1%7D%7Bx%7D%20)
This implies that;
![\frac{1}{c} = \frac{ ln(7) - 0}{6}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7Bc%7D%20%3D%20%5Cfrac%7B%20ln%287%29%20-%200%7D%7B6%7D%20)
![\frac{1}{c} = \frac{ ln(7)}{6}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bc%7D%20%3D%20%5Cfrac%7B%20ln%287%29%7D%7B6%7D%20)
![c = \frac{6}{ ln(7) }](https://tex.z-dn.net/?f=c%20%3D%20%5Cfrac%7B6%7D%7B%20ln%287%29%20%7D%20)
![c\approx 3.08](https://tex.z-dn.net/?f=c%5Capprox%203.08)
Since the function is continuous on [1, 7] and differentiable on (1, 7) it satisfies the mean value theorem.
X = -4y+3
-x-4y = -3
________________
x+4y = 3
-x-4y = -3
0 = 0
Possible and determined system (single solution)
6 because the 36 is r² where r is the radius, r² = 36; we have to isolate the r so it's by its self, so square root both sides, r = 6
Answer: The answer is C) y= 2/3x-3
Step-by-step explanation:
i figure it out